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(A) The resistance of the original wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r_1^2}$,where $r_1 = 9 \ mm$.Given $R = 5 \ \Omega$,so $

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$7.5$

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(A) The resistance of the original wire is given by $R = ho \frac{l}{A} = ho \frac{l}{\pi r_1^2}$,where $r_1 = 9 \ mm$.Given $R = 5 \ \Omega$,so $5 = ho \frac{l}{\pi (9 	imes 10^{-3})^2} \quad \dots (i)$When replaced by $6$ wires of radius $r_2 = 3 \ mm$ connected in parallel,the resistance of each new wire is $R' = ho \frac{l}{\pi r_2^2} = ho \frac{l}{\pi (3 	imes 10^{-3})^2}$.Since the $6$ wires are in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{6}{R'}$.Thus,$R_{eq} = \frac{R'}{6} = \frac{1}{6} \left( ho \frac{l}{\pi (3 	imes 10^{-3})^2} ight) \quad \dots (ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{R}{R_{eq}} = \frac{ho l / (\pi 	imes 81 	imes 10^{-6})}{ho l / (6 	imes \pi 	imes 9 	imes 10^{-6})} = \frac{6 	imes 9}{81} = \frac{54}{81} = \frac{2}{3}$.Therefore,$R_{eq} = \frac{3}{2} R = \frac{3}{2} 	imes 5 = 7.5 \ \Omega$.

An electrical cable consists of a single copper wire of radius $9 \ mm$. Its resistance is $5 \ \Omega$. If this cable is replaced by $6$ copper wires,each of radius $3 \ mm$,then the total resistance of the cable will be ............. $\Omega$.

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